Hi everyone and welcome. We're going to talk about the fundamental theorem of calculus, part one of two. But before we do so, let's just talk about what's to come. This theorem, the fundamental theorem of calculus, as you can imagine by its name, is pretty darn important, and let me just tell you why, we're going to talk about it and where it's going. Remember so far we have the definite integral, and we're studying accumulation. We have approximations to do this, but oftentimes we want the exact accumulation over a certain interval, and the graphs that we've seen, they tend to have nice shapes like maybe it's a rectangle, maybe it's a triangle, maybe something nice that we know, trapezoid or something, but most, functions don't fall into this special category of having these nice shapes. What's going to happen in the videos in sections to come, we're going to start to develop methods, to calculate definite integrals of functions that are not in this category. They don't have shapes. The way to do that is mildly amazing. It is going to use differentiation to do this. It turns out that integration and differentiation are just related to each other. They are inverse operations to say weighing like addition and subtraction or e and log, that they undo each other. This relationship, is what the fundamental theorem of calculus which is going to put down and be specific about, and it allows us to compute definite integrals, by completely reframing the question, and putting it in terms of differentiation, and that's why it's called the fundamental theorem of calculus. Let's get started with that first, and let me do that by putting a definition down, so we say an antiderivative of a given function is some function F of x, such that the derivative of the function is equal to f, so this is called an antiderivative, and you can imagine that it's like the Jeopardy of function. So I'll give you the answer and you give me the question. So the ideas are given some little function F, then what function would have, the derivative of what function, shows what function is equal to F of x, this thing. Just to give you some examples here. Let start of relatively easy. If I start off with the right function, let's say 4x cubed, I think this is the answer, this is the Jeopardy question. What function was I originally thinking about? What function has derivative 4x cubed? So I think a little bit to say, that's just x_4, and I'm wrong here you got also have x to the fourth, or you can have more than one answer, x_4 plus five, or x to the fourth plus six or whatever, and in general, when we write the general antiderivative, we'll say x_4 plus c, and c will be some constant. Normally will last for is like finding an antiderivative, this is a little different here because derivatives are unique, but antiderivatives, there are a whole family of functions as we say and you can do this. We're going to use antiderivatives in what is to follow, and remember that if you find one of them, like x_4, you found them all, just put on the plus c and you get all of them. We are now going to define a new function, to do that, let's set it up first. Let me do in the next slide, so we have a clean slide. Now we're going to define a new function, and we are going to start with some function f of x, a function, and we'll start a inside the domain of f. Pick a function, pick an element in its domain, then define, I'll call it A of t, A for accumulation. I'm going to find new function, and it's going to be unlike anything you've seen before, it's defined by fixing the base, so this given value A, and let t be the variable, and it's going to be the definite integral from a to t, of f of x dx. Now there's a lot of letters going on here, but we'll talk about in a second, but this function is called the net accumulation function. This is the net inside of the slide here, accumulation function. So stare at this for a second and let this thing sink in. What is this function? How does it work? It's very different than anything else that you've seen before and you have to keep in mind that this is a function. One input has one output. Let's do a generic thing here. What's given is the function. The function is given, f of x is some given function, and the lower limit A is also given. All these things are given. There's already a function given, this is why it looks confusing because here's another function. We're going to build another function out of this thing and let's give you a specific value. Let's say like t_1, some point in time. The way this works is you plug in t_1 to the function and then the integral is calculated the definite integral from a to this value of the function, so it accumulates the area under the function. Whatever this number is, remember this is the number, whatever the area is under the function from A to t_1, that is what the function kicks back, that number. If you vary the variable as they tend to get varied, so let's draw it again. Let's put A on the map, let's pick a different value. Use different color. Let's put over here t_2, you can plug this number into the formula and it will compute a different number, it has accumulated more net area and this number would be the net area under the curve above the x axis, in this particular case, from A to t_2, and that's some number. Number goes in, number comes out. You can do this all day and that's why it's a nice little function. In terms of what number to pick for A or what function affect this, they can vary. Depending on what you're trying to do, those can vary, it doesn't matter, but once you pick a function and a value A, then you have this net accumulation function that you can do. Let's do an actual example with some specific numbers just to get used to this thing. As an example, let's pick a equals negative one, F of x is the constant function six. Let's draw a picture for a minute. I have negative one on the board, I have the positive number six. Just map it there, picture not drawn to scale and so far so good, those are given. Now I want the accumulator function, which is built off this thing, we're going to start at a lower bound of minus one. We're going to go all the way to t and I put in my function six dx. All I want to do when presented with new function, think about that day when you learn what function was, is I just want to evaluate this at certain values. I just want to take A of negative one. What's that? I also want to do A of positive one and let's also evaluate this at A of five. It's amazing because we've gone back to the very beginning of functions. How do we evaluate a function? But this function is complicated. It's defined using an integrals, we have to be very careful here. When I plug in negative one, this function asked for the net accumulation from minus 1-1 of six dx. This is how much has been accumulated. If the bounds on our definite integral are the same, I've accumulated nothing. The total area under this thing is nothing, so this is zero. So this function a of negative one is zero. Let's actually get a non-zero value. If I go all the way to one, this function is asking what net area have you accumulated from minus one to one of the function six. The picture that would go along with this one is how much areas under the graph from minus one to one. I pick this example, you can imagine the basis is a nice little rectangle base is two, the height is six, so six and two. This turns out to be then total accumulated area that is just 2 times 6, which is 12. I plug in one, I get back 12. Similar with five, let's just switch it up for colors for a second, if I go all the way to five, I want the total net accumulated area under the graph from my given the lower limit of minus one to five, six dx. Now that got a little bigger, but again, we can still work this out. The base in this case is six, the height is six area of rectangle, 6 times 6, 36. This is how this function works. The function will be given, you could do it for more complicated functions may be rectangles and triangles. But we can do these things. This is this net accumulator function and the fundamental theorem of calculus part 1, the first fun one, talks about specifically this accumulator function. Now without further due, I present to you the statement of the theorem. Here it is. If I have some function f of x, and it's a nice smooth function, so a smooth function, so it has derivative, nice smooth function, and I pick some number a inside the domain, get over there, the domain of the function to pick some element, I think. Then, and here it comes, I'll put it in words first and see if we can convert it. Then the net accumulator function a of t, defined by the integral from a to t of f of x dx is an anti-derivative, it's an anti-derivative of the function. I'm going to say it again, give it a function and, then on the domain, you can find an anti-derivative. Let me put that in symbols for you. That says that if I take the derivative with respect to, remember the variable is t, x is given, so the derivative with respect to t of the accumulator function, let me just put the accumulator function down in its integral form, is equal to f of t. You can see the derivative and the integral playing a role together here. Like, and again, the adjective fundamental in the title of this theorem, it reflects the deep significance of this result. It bridges and interconnects differentiation and integration. Like you can't get more beautiful than that. Just the strictly computational level, it just allows us to provide another derivative formula for a new kind of function. This also shows that, by the way, if I have any smooth function, I will always have an anti-derivative, which is nice. That's good. Let's just actually do one of these, will do more of these of course, but let's do one with specific examples just on a computational level. Let's find the derivative of the accumulator function. Let's go from 2 to t of x_fourth, plus 2_x squared plus 1 dx. I have some quartic function going from 2 to t. How do I do this? Use the first fundamental theorem of calculus, and it says the derivative of the accumulation function is the original function with a variable t. This is just good old t_fourth, 2_t squared plus 1. This derivative is, put that onto the list of easy ones, although it has fancy notation, but it's just as easy as e_x or something like that. Speaking of which, why not? Let's do one more just for fancies. Did I just say fancies? Yes I did. D dt of, let's go nuts here, t to 1, e_x cubed dx. Now you can see I'm turned little curve ball in here, I put the variable downstairs. What do we do with that? Should we panic? No we should not. We know how to handle that. I can flip the bounds. Do you remember the costs though? You can flip the bounds but at a cost of a minus sign, so flip it 1 to t, e_x cubed dx. Now it's a derivative. Can minus signs like constant times minus 1. That just pops right out, so you can think of that as negative d dx over the integral from 1 to t of e_x cubed dx. Now we're back in familiar territory. I can use the first fundamental theorem of calculus and this becomes negative, don't forget the negative sign, e_t cubed. Very, very good. Let this thing sink in, I appreciate that we have a new function and a new derivative, and of course, where I tried to explain the interplay of the derivative and integrals, and we'll do more examples and videos to come. Good job on this one, and see you next time.
